在最早学习四则运算的过程中，我们其实就已经掌握了进制算法，这一次我将对二进制运用这个进制算法来实现四则运算。
四则运算
math.c
/*** 功能：通过位操作实现四则运算* 算法：完全参照十进制的进制算法** Created with CLion* User: zzzz76* Date: 2018-02-10*/#include <stdio.h>
#include <assert.h>
#include "math.h"/*** 加法：往上递归实现** @param a* @param b* @return*/
int base_add(int a, int b) {/* 值在函数中的传递只能通过参数或者返回操作，所以递归效果无非是体现在参数和返回操作上 */if (b == 0) {return a;}int save = a ^b;int promote = (a & b) << 1;return base_add(save, promote);
}/*** 加法：迭代实现** @param a* @param b* @return*/
int base_add_re(int a, int b) {while (a && b) {int promote = (a & b) << 1;a = a ^ b;b = promote;}return a ^ b;
}/*** 减法：往上递归实现** @param a* @param b* @return*/
int base_sub(int a, int b) {if (b == 0) {return a;}int save = a ^ b;int reduce = ((~a) & b) << 1;return base_sub(save, reduce);
}/*** 减法：迭代实现** @param a* @param b* @return*/
int base_sub_re(int a, int b) {while(b) {int save = a ^ b;b = ((~a) & b) << 1;a = save;}return a;
}/*** 减法：补位实现** @param a* @param b* @return*/
int base_sub_re_re(int a, int b) {return base_add(a, base_add(~b, 1));
}/*** 乘法：递归实现** @param a* @param b* @return*/
int base_mul(int a, int b) {int count = 0;if (a == 0) {return count;}if (a & 1) {count = base_add(count, b);}a = (unsigned)a >> 1;b <<= 1;count = base_add(count, base_mul(a, b));return count;
}/*** 乘法：迭代实现** @param a* @param b* @return*/
int base_mul_re(int a, int b) {int count = 0;while (a) {if (a & 1) {count = base_add(count, b);}a = (unsigned)a >> 1;b <<= 1;}return count;
}/*** 除法：迭代实现** @param a* @param b* @return*/
int base_div(int a, int b) {assert(b);int result = 0;int bit_num = 31;while (bit_num != -1) {if (b <= ((unsigned) a >> bit_num)) {result = base_add(result, 1 << bit_num);a = base_sub(a, b << bit_num);}bit_num = base_sub(bit_num, 1);}return result;
}/*** 除法：递归实现** @param a* @param b* @return*/
int base_div_re(int a, int b) {assert(b);if (a < (unsigned) b) {return 0;}int bit_num = 0;while (b <= (unsigned) a >> 1 >> bit_num) {bit_num = base_add(bit_num, 1);}int result = 1 << bit_num;a = base_sub(a, b << bit_num);result = base_add(result, base_div_re(a, b));return result;
}
math.h
/*** Created with CLion* User: zzzz76* Date: 2018-02-12*/#ifndef MATH_H
#define MATH_Hint base_add(int a, int b);
int base_add_re(int a, int b);int base_sub(int a, int b);
int base_sub_re(int a, int b);
int base_sub_re_re(int a, int b);int base_mul(int a, int b);
int base_mul_re(int a, int b);int base_div(int a, int b);
int base_div_re(int a, int b);#endif //MATH_H
test.c
/*** Created with CLion* User: zzzz76* Date: 2018-02-12*/#include "math.h"
#include <stdio.h>static int main_ret = 0;
static int test_count = 0;
static int test_pass = 0;#define EXPECT_EQ_BASE(expect, actual, format) \do {\test_count++;\if ((expect) == (actual)) {\test_pass++;\} else {\fprintf(stderr, "%s:%d: expect: " format " actual: " format "\n", __FILE__, __LINE__, expect, actual);\main_ret = 1;\}\} while(0);#define EXPECT_EQ_INT(expect, actual) EXPECT_EQ_BASE(expect, actual, "%d");#define TEST_BASE_ADD(expect, a, b) \EXPECT_EQ_INT(expect, base_add(a, b));\EXPECT_EQ_INT(expect, base_add_re(a, b));
static void test_base_add() {TEST_BASE_ADD(2, 1, 1);TEST_BASE_ADD(0, -1, 1);TEST_BASE_ADD(0, 1, -1);TEST_BASE_ADD(-2, -1, -1);TEST_BASE_ADD(-2147483648, 2147483647, 1);
}#define TEST_BASE_SUB(expect, a, b) \EXPECT_EQ_INT(expect, base_sub(a, b));\EXPECT_EQ_INT(expect, base_sub_re(a, b));\EXPECT_EQ_INT(expect, base_sub_re_re(a, b));
static void test_base_sub() {TEST_BASE_SUB(0, 1, 1);TEST_BASE_SUB(-2, -1, 1);TEST_BASE_SUB(2, 1, -1);TEST_BASE_SUB(0, -1, -1);TEST_BASE_SUB(2147483647, -2147483648, 1);
}#define TEST_BASE_MUL(expect, a, b) \EXPECT_EQ_INT(expect, base_mul(a, b));\EXPECT_EQ_INT(expect, base_mul_re(a, b));
static void test_base_mul() {TEST_BASE_MUL(9, 3, 3);TEST_BASE_MUL(-9, -3, 3);TEST_BASE_MUL(-9, 3, -3);TEST_BASE_MUL(9, -3, -3);TEST_BASE_MUL(0, -2147483648, 2);TEST_BASE_MUL(-2, 2147483647, 2);
}#define TEST_BASE_DIV(expect, a, b) \EXPECT_EQ_INT(expect, base_div(a, b));\EXPECT_EQ_INT(expect, base_div_re(a, b));
static void test_base_div() {TEST_BASE_DIV(2, 2, 1);TEST_BASE_DIV(-2, -2, 1);TEST_BASE_DIV(0, 2, -1);TEST_BASE_DIV(0, -2, -1);TEST_BASE_DIV(0, 1, 2);TEST_BASE_DIV(0, 1, -2);TEST_BASE_DIV(2147483647, -1, 2);TEST_BASE_DIV(1, -1, -2);
}static void test_base() {test_base_add();test_base_sub();test_base_mul();test_base_div();
}int main() {test_base();printf("%d/%d (%3.2f%%) passed\n", test_pass, test_count, test_pass * 100.0 / test_count);return main_ret;
}
从递归角度看待代码
递归是函数调用，考虑值的传递过程，适合阅读。
迭代是具体的实现过程，往往代码效率更加充分。
通常我们在写代码时，往往注重代码的效率和正确性，而忽略了代码表达的意思，致使代码难以阅读。所以对两者的取舍，一定程度影响了代码的好与坏
